Solubility Product Constant (Ksp) - Online Tutor, Practice Problems & Exam Prep

Solubility, a key chemical property, refers to a solute's ability to dissolve in a solvent, quantified by the solubility product constant (K_sp). A higher K_sp indicates greater solubility of ionic solids. The dissolution process involves breaking ionic solids into their respective ions, leading to an equilibrium expression that excludes solids. K_sp can be calculated from known solubility values, while solubility can also be derived from known K_sp values, emphasizing the interdependence of these concepts in understanding ionic compounds.

Solubility Product Constant (Ksp) Concept 1

Now recall that solubility is a chemical property that deals with the ability of a solute to become dissolved in a solvent. Now connected to this idea of solubility is a new term, our solubility product constant, which uses the variable ksp. Now this is the equilibrium constant that deals with the solubility of ionic solids. So if we're looking at the solubility of ionic solids, we're dealing with ksp. We're going to say here, it deals with their solubility, and we're going to say solubility can also be referred to as concentration or molarity. Remember molarity uses capital M. Now the basic idea is the higher your ksp value, then the more soluble your ionic solid. And the smaller your ksp value then the less soluble your ionic solid, the less of it will dissolve within a solvent. So just keep this in mind when looking at Ksp.

Solubility Product Constant (Ksp) Example 1

It asks, which substance is the most soluble? So here we have silver chloride, magnesium carbonate, calcium sulfate, and copper(II) sulfide, each of them with their given K_sp value. Remember, we want the most soluble of all of these choices. Remember, we're going to say the greater your K_sp value, then the more soluble you are as an ionic solid. So if we look here, we'd say that the answer has to be C, since it's 7.1 × 10^-5. It has the smallest negative, so it's the largest value there. It would be the most soluble out of all four choices. So here, option C would be the correct answer.

Solubility Product Constant (Ksp) Concept 2

Now, placing an ionic solid within a solvent involves 2 competing processes. We have the dissolution reaction, so basically where it's being broken up, and we also have its reverse. Okay? So dissolution looking at the forward direction of the reaction where ionic solid dissolves into its ions. So here we have a b 3 a 2 solid. So if we were to break this up into its ions, we'd say we have 2a superscript 3 plus ⁺, ions in solution are aqueous, plus 3b superscript 2 minus ⁻ aqueous. And remember, we know that the first ion is positive and the second one is negative because that's the order in which we write ionic compounds. Now we're going to say that from this equilibrium equation, its equilibrium expression can be determined. Now the equilibrium expression is just the ratio of concentrations of products over reactants. And we're gonna say that recall the equilibrium expression ignores solids and liquids for its ratios. Right? So it's important that we know how to break up our ionic solids into their respective ions in order to successfully write your equilibrium expression later on. So keep this in mind in terms of the parameters in which it operates. It's products over reactants, and it ignores solids and liquids.

Solubility Product Constant (Ksp) Example 2

Here we need to provide the equilibrium expression for calcium nitrate. So step 1 says we need to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Remember, in solution, our ions are in the aqueous form. Alright. So, we have calcium nitrate solid. It breaks up into calcium, which is in group 2A, so it's 2+. That's where this 2 came from. Not only is that where the 2 came from, but that also means that we have 2 nitrate ions. So that's 2 NO₃^-, remember the charge of nitrate ion, aqueous. Now that we've done this, we're going to say step 2, using K_sp, write the equilibrium expression based on the equilibrium equation. Since the reactant is a solid, set it equal to 1 within the equilibrium expression. Remember, it's products over reactants. So we'd have calcium in brackets, so 2 + count, Ca²⁺ times NO₃^-. And remember this 2 here, whatever the coefficient is that becomes the power. So this would be a 2 here. Our reacting as a solid, so we just ignore it and replace it with 1. But realize here that we have this expression over 1, which just translates into those concentrations times each other. And nitrates would be squared. Okay. So that would just be our equilibrium expression for this calcium nitrate solid. So, these are the steps you need to take in breaking up your ionic solid into its respective ions, and then correctly giving the equilibrium expression for it.

Ca(NO 3 ) 2 → Ca 2+ (aq) + 2 NO 3 - (aq) K sp = [Ca 2+ (aq)] ∗[NO 3 - (aq)] 2

Solubility Product Constant (Ksp) Concept 3

We can say if the solubility or molar concentrations of ions within ionic solids are known, the K_sp can be calculated. If we take a look here, we say calculate the K_sp value for silver phosphate, which is Ag₃PO₄, which has a solubility of \(1.8 \times 10^{-18}\) at 25 degrees Celsius. Here we say write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Alright. So here we have our silver phosphate solid. It's going to break up into its ions. It's going to break up into 3 silver ions, remember ions are aqueous in solution, plus our phosphate ion aqueous.

Next, we write the equilibrium expression based on the equilibrium equation. So remember, equilibrium expression is K_sp equals products over reactants. But your reactant is a solid, so we're going to ignore it. Thus, it's just going to be Ag⁺ cubed, remember the coefficient here is going to become the power, so cubed, times PO₄³⁻. Now some new steps. Step 3, we're going to make concentrations of the ions equal to their coefficients multiplied by the x variable. Alright. So what do I mean by that? We're going to say this is equal to the coefficient in front of the silver ion in our equation which is a 3, so this equals 3x and it's going to be cubed. And there's a coefficient of 1 here that's invisible, so that would be times 1x or just x.

Next, we are going to substitute in the given solubility value for the x variable and solve for K_sp. Before we do that, we have to work out this algebraic expression here. So \(3x\) cubed means that 3 is going to be cubed and x is going to be cubed. 3 times 3 times 3 equals \(27 x^3\) times x. So \(27 x^3\) times x is \(27 x^4\). So that is what my K_sp is equal to. So now we're going to do what step 4 said. Substitute in the given solubility value; we're told the solubility is this \(1.8 \times 10^{-18}\). So K_sp equals \(27 \times (1.8 \times 10^{-18})^4\). Order of operations. We do what's in the parentheses first. So \(1.8 \times 10^{-18}\) to the 4th. When you punch that into your calculator, it's going to give you back \(1.04976 \times 10^{-71}\). Then that gets multiplied by 27. When we do that, we get as our answer \(2.8 \times 10^{-70}\) as the K_sp for our silver phosphate here. Here our answer has 2 significant figures because the solubility given to us at the beginning also has 2 significant figures.

Solubility Product Constant (Ksp) Concept 4

Conversely, the solubility of an ionic solid can be determined when its K_sp value is already known. So here it says the K_sp value for strontium fluoride SrF₂ is 7.9 times 10 to the negative 10 at 25 degrees Celsius. Calculate its solubility in molarity. Right. So step 1, we're going to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. So that part doesn't change. We have strontium fluoride solid which breaks up into a strontium ion plus 2 fluoride ions.

Next, we write the equilibrium expression based on the equilibrium equation. So that's K_sp equals our strontium ion times our fluoride ion squared. And remember the coefficient becomes the power. Next, we solve for the solubility variable x based on the given K_sp value. Alright. So remember, to do that we're going to say, we have our K_sp here, which is 7.9 times 10 to the -10, and that equals x, which stands in for the strontium ion, times, remember the coefficient is going to play a part here, it's going to be 2 times x, and it's still squared. So we're going to have 7.9 times 10 to the negative 10 equals x times, 2 squared is 2 times 2, so that's 4, x squared. So we're just solving this like a math problem. So it's going to be 7.9 times 10 to the negative 10 equals 4 x³. We're going to divide both sides now by 4 in order to isolate our x³, so when we do that we're going to get here x³ equals 1.975 times 10 to the negative 10.

And here we need to take the cube root of both sides in order to just isolate my x here. So we're going to say here, when we take the cube root of both sides, x equals at the end 5.8 times 10 to the negative 4 molar as my final concentration for this strontium fluoride solid.

Determine the equilibrium expression of the barium nitride solid.

• A.
• B.
• C.
• D.

Manganese (V) hydroxide has a measured solubility of 3.4×10^–5 M at 25ºC. Calculate its K_sp value.

• A. 5.7×10^–9
• B. 1.5×10^–27
• C. 7.7×10^–27
• D. 4.8×10^–24

The K_sp value for strontium fluoride, SrF₂, is 7.9×10^–10 at 25ºC. Calculate its solubility in g/L.

• A. 5.8×10^–4 g/L
• B. 7.3x10^-2 g/L
• C. 4.6×10^–6 g/L
• D. 0.092 g/L